The Relativity of Simultaneity

[arfa brane]

Motor Daddy: there are two frames of reference: the embankment, and the train.

Is it really that difficult to understand?

Pete said

The train observer measures the train to be 10.000000001367545 rulers long.

Where the train observer is on the embankment, and "at rest". So the observer, who is not on the train, sees different lengths than an observer who is on the train and moving with it.

Yeesh.

 

[Tach]

So the light travel time one way is different than the light travel time the other way, when measured on the train?

Not on the train, in the embankment frame. How dense are you?

 

[Motor Daddy]

Motor Daddy: there are two frames of reference: the embankment, and the train.

Is it really that difficult to understand?

Pete said
Where the train observer is on the embankment, and "at rest". So the observer, who is not on the train, sees different lengths than an observer who is on the train and moving with it.

Yeesh.

It simply boils down to the fact that the train observer and the embankment observer both agree that the strikes occurred at A and B simultaneously.

I say the strikes occurred at A and B simultaneously and that both observers agree, and Pete (and Einstein) says the observers don't agree with each other as to when the strikes occurred at A and B.

I showed mathematically that the strikes occurred at exactly 12:00:00 at A and B, and that both observers agree with that. Pete says I'm full of it and he can prove it. I'm all ears!! First we have to settle on the details, and so far, there is a thorn in Pete's side, and that is that he has acknowledged that it takes light a different amount of time to travel the length of the train, depending on which direction you measure it. Exactly my point, and how I am able to calculate the absolute velocity from within.

We'll see.......

 

[Motor Daddy]

Not on the train, in the embankment frame. How dense are you?

The stick is on the train, and these are times he gave me.

From front to back, it takes t=d/(c+v) = 3.335585787248 ns
From back to front, it takes t=d/(c-v) = 3.335696117627 ns

Are you speaking for Pete now and going to give me different times from both the train and the embankment of how much time it takes for light to travel the length of the stick when it is on the train? The actual distance and times are irrelevant, the point is that it takes a different amount of time for light to travel the length of the stick, depending on which way you measure it, as light has to travel further in one direction than the other. Prove it doesn't!

 

[Tach]

The stick is on the train, and these are times he gave me.

You are totally dense. No chance of ever getting well.

the point is that it takes a different amount of time for light to travel the length of the stick, depending on which way you measure it, as light has to travel further in one direction than the other. Prove it doesn't!

That was precisely the point Pete made to you and you just denied it because you didn't understand it.

 

[Motor Daddy]

You are totally dense. No chance of ever getting well.

That's what I though.

You are among the camp that says that it takes the same time for light to travel from a lamp post to a person at 10 meters, a person at 15 meters, and a person at 20 meters.

Point being, If I am a distance away from a lamp post and measure the amount of time it takes for light to reach me, you say it takes the same amount of time for the light to reach me if I start running away from the lamp as soon as the light departs the lamp. You're clueless!

...and to put icing on the cake, you say the time is the same if I start running towards the lamp when the light is emitted.

 

[Tach]

That's what I though.

You are among the camp that says that it takes the same time for light to travel from a lamp post to a person at 10 meters, a person at 15 meters, and a person at 20 meters.

Not at all. I am in the camp that has determined that you are totally dense and that you can't follow the simplest proofs because you can only engage mouth and not brain.

 

[arfa brane]

The stick is on the train, and these are times he gave me.

The stick is on the train, the observer is not on the train but standing on the "stationary" embankment.

Oh, and the stick is not an observer, just in case that needed clarification.

 

[Motor Daddy]

The stick is on the train, the observer is not on the train but standing on the "stationary" embankment.

Oh, and the stick is not an observer, just in case that needed clarification.

Just to clarify, each end of the stick IS an observer, with a clock!

 

[Tach]

Just to clarify, each end of the stick IS an observer, with a clock!

Yep, you have the brain of a stick.

 

[Pete]

Pete said
Where the train observer is on the embankment, and "at rest". So the observer, who is not on the train, sees different lengths than an observer who is on the train and moving with it.

No.
The train observer is riding on the train.

Not at all. I am in the camp that has determined that you are totally dense and that you can't follow the simplest proofs because you can only engage mouth and not brain.

Tach, you're not helping.

Guys, please let me work through this exercise with MD without interjections. Thanks.

 

[Pete]

What do you mean by "the actual travel times?"

The time measured using clocks we know are synchronized and not dilated, ie the embankment clocks.

Do you plan to use a different sync method from the one you (indirectly) agreed to using for the embankment?

Not unless you want to. We'll get to synchronizing train clocks shortly. One step at a time, right?

The clocks on the embankment are absolutely sync'd.

Yes, that's the assumption I'm working with.

 

[Motor Daddy]

We'll duke it out in a moment when it comes to the train clocks, but you may proceed, I'm good with the embankment.

 

[arfa brane]

No.
The train observer is riding on the train.

Oops, my bad.

However, since there are clocks on the embankment which are synchronous, the observer on the train is able to postulate a "virtual" observer on the embankment, i.e. the clocks. This is because the observer on the train can see the same clocks, and mentally "transform" their location to the stationary embankment.

Hope I'm not confused about that.

 

[Pete]

Recap - the story so far:

Assumptions:

• The embankment is at rest
• Light travels at c with respect to the embankment
• Clocks on the embankment are synchronized with each other
• The train observer knows that light travels at c with respect to something at rest
• The train observer doesn't know that the embankment is at rest
• The train observer doesn't know that the embankment clocks are synchronized
• The train observer has precise clocks, but he doesn't know if they're synchronized
• Moving clocks run slowly by the Lorentz factor
• Moving rulers are shorter in the direction of motion by the Lorentz factor

The scenario

Point A and point B are marked 10 metres apart on the embankment.
Point A' is moving, marked on the back of the train.
Point B' is moving, marked on the front of the train.
An observer M is standing on the embankment, halfway between point A and point B.
An observer M' is standing on the train, halfway between point A' and point B'.

The train passes the embankment at 4,958 m/s
gamma = 1 / sqrt(1-v^2/c^2) = 1.0000000001367545054905367903816

At t=0.000:

• the front of the train is passing point B
• the back of the train is passing point A
• the train observer M' is passing embankment observer M
• M' has a clock with him that reads t'=0.000
• A bolt of lightning strikes the front of the train and point B
• Another bolt of lightning strikes the back of the train and point A

From this, I conclude that:

• the lightning bolts struck simultaneously
• the moving train is 10 actual metres long.
• rulers on the train are contracted to 1/gamma = 0.99999999986325 actual metres long
• The train observer measures the train to be 10.000000001367545 rulers long
• clocks on the train are dilated, elapsing 1/gamma = 0.99999999986325 seconds every actual second
• It takes light t=d/(c+v) = 3.335585787248 ns to go from the front of a train ruler to the back of a train ruler
• It takes light t=d/(c-v) = 3.335696117627 ns to go from the back of a train ruler to the front of a train ruler

Now, the next step...

At t = d/(c+v) = 16.67792893852027 ns :

• The flash from lightning bolt B reaches M' (the train observer)
• The clock at M' reads t' = t/gamma = 16.67792893623949 ns

At t = d/c = 16.67820475990760 ns:

• The flash from both lightning bolts reaches M

At t = d/(c-v) = 16.67848059041821 ns:

• The flash from lightning bolt A reaches M'
• The clock at M' reads t' = t/gamma = 16.67848058813736 ns

What can the train observer conclude from the readings on his clock?
Only that his clock elapsed 0.55 picoseconds between the front lightning flash reaching him and the rear flash reaching him.
But since doesn't know how fast he's going in what direction, and there are no clocks anywhere else on the train that are synchronized with the centre clock, he can't tell from that data alone which flash occurred first.

Questions or problems?

So how will we proceed?
Sync clocks first, or measure velocity without synced clocks?

• We could have M' measure his velocity using a single clock, a light flash, and a mirror
• We could have M' measure his velocity using unsynchronized clocks (flash one way, flash the other way)
• We could have M' synchronize the clocks at each end of the train with each other in some way. Perhaps we could send identical robots from the centre to each end, starting in the middle at the same

Which way would you prefer?
Shall we try all of them?
Other suggestions are welcome.

 

[Pete]

Oops, my bad.

However, since there are clocks on the embankment which are synchronous, the observer on the train is able to postulate a "virtual" observer on the embankment, i.e. the clocks. This is because the observer on the train can see the same clocks, and mentally "transform" their location to the stationary embankment.

Hope I'm not confused about that.

In this scenario, we're not (yet) allowing the train observer to see the embankment clocks.

All the train observer has is his contracted rulers, his dilated clock/s, and the flashes of lightning.

 

[Motor Daddy]

The light travel times are correct, the clock readings on the train are incorrect. You have not established that the clocks on the train tick any faster or slower than the embankment's clocks. Faster or slower than what? As far as the train observer is concerned, his clocks are the ONLY clocks in the universe. He has no windows to the outside world.

I agree with your light travel times, they are exactly the same as my travel times in my example.

The embankment observer had both lights hit him simultaneously.
The embankment is at a zero velocity.
Light traveled equal distance to reach the embankment observer.

What time was it on the embankment observer's watch when the lights hit him?

 

[Motor Daddy]

Also, just to stay in sync with Einstein's Chapter 9, A is at the rear of the train and B is at the front of the train, so you have the times reversed in your post. I'm sure that's what you meant, it's just a small technicality.

 

[Motor Daddy]

...and just a side note:

You've just proven that in the train frame, it takes more time for light to travel from A to the observer at the midpoint than it does for light to travel from B to the observer at the midpoint.

In the train frame, those are equal distances, and yet light takes different amounts of time to travel that same distance.

That is a fact, so you have some explaining to do.

 

[Pete]

The light travel times are correct, the clock readings on the train are incorrect. You have not established that the clocks on the train tick any faster or slower than the embankment's clocks. Faster or slower than what? As far as the train observer is concerned, his clocks are the ONLY clocks in the universe. He has no windows to the outside world.

The train observer doesn't yet know if his clocks are dilated or by how much, but we do.

We've established that the embankment is at rest.
We've established that the train is moving at 4958m/s.
We've established that in the mathematical world we're working in, moving clocks run slowly.
So we conclude the train clocks are dilated, elapsing one tick every 1.0000000001367545 seconds.

What time was it on the embankment observer's watch when the lights hit him?

The flashes reached him at t=16.67820475990760 ns, so that's what the embankment clock (which we've said are at rest, undilated, and absolutely sync'd) at M reads.

Also, just to stay in sync with Einstein's Chapter 9, A is at the rear of the train and B is at the front of the train, so you have the times reversed in your post. I'm sure that's what you meant, it's just a small technicality.

I'll edit the previous posts to match that convention.

...and just a side note:

You've just proven that in the train frame, it takes more time for light to travel from A to the observer at the midpoint than it does for light to travel from B to the observer at the midpoint.

In the train frame, those are equal distances, and yet light takes different amounts of time to travel that same distance.

That is a fact, so you have some explaining to do.

No, the light travel times haven't yet been measured by the train observer.
All he knows is what his clock read when each flash reached him.
He doesn't know when the lightning actually struck, because he doesn't have synchronized clocks at the ends of the train.
We know that they struck simultaneously, but the train observer doesn't.

Are you ready for the train observer to synchronize his clocks, or to measure his velocity?